Question: $g(x)=(x-12)^2-30.25$ 1) What are the zeros of the function? Write the smaller $x$ first, and the larger $x$ second. $\text{smaller }x=$
Solution: $\begin{aligned} (x-12)^2-30.25&=0 \\\\ (x-12)^2&=30.25 \\\\ \sqrt{(x-12)^2}&=\sqrt{30.25} \\\\ x-12&=\pm 5.5 \\\\ x&=\pm5.5+12 \\\\ x={6.5}&\text{ or }x={17.5} \end{aligned}$ $g(x)$ is given in vertex form: $g(x)=(x-{12})^2{-30.25}$ So the vertex of the parabola is at $({12},{-30.25})$. In conclusion, $\begin{aligned} \text{smaller }x&=6.5 \\\\ \text{larger }x&=17.5 \end{aligned}$ The vertex of the parabola is at $(12,-30.25)$